Let $g(x)=-2x^4+x^2-3x$. $g'(x)=$
Explanation: According to the sum rule, the derivative of $-2x^4+x^2-3x$ is the sum of the derivatives of $-2x^4$, $x^2$, and $-3x$. The derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ For example, this is the derivative of the first term: $\begin{aligned}\dfrac{d}{dx}(-2x^4)&=-2\dfrac{d}{dx}(x^4)&&\gray{\text{Constant multiple rule}}\\\\ &=-2\cdot (4x^3)&&\gray{\text{Power rule}}\\ \\ &=-8x^3\end{aligned}$ Here is the complete differentiation process: $\begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}(-2x^4+x^2-3x) \\\\ &=-2\dfrac{d}{dx}(x^4)+\dfrac{d}{dx}(x^2)-3\dfrac{d}{dx}(x)&&\gray{\text{Basic differentiation rules}} \\\\ &=-2\cdot 4x^3+2x-3\cdot 1x^0&&\gray{\text{The power rule}} \\\\ &=-8x^3+2x-3 \end{aligned}$ In conclusion, $g'(x)=-8x^3+2x-3$.